What is off-by-one in the index?

The nth term uses (n - 1), not n, as the exponent or multiplier. The first term corresponds to n = 1, so after k steps you are at term k + 1.

QLDGeneral MathematicsSyllabus dot point

Topic 2: Growth and decay in sequences - how do we model quantities that change by a constant amount or a constant ratio?

Use arithmetic and geometric sequences and first-order recurrence relations to model linear growth or decay and geometric growth or decay, find the nth term and partial sums, and apply these models to simple interest, reducing-balance depreciation and compound contexts

A focused answer to the QCE General Mathematics Unit 3 dot point on sequences and change. Covers arithmetic and geometric sequences, first-order recurrence relations, nth-term and sum formulas, and modelling linear versus geometric growth and decay including simple interest and depreciation, with arithmetic-verified worked examples for IA2 and the external assessment.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

QCAA wants you to model change with sequences: a constant amount added each step (arithmetic, linear growth or decay) or a constant ratio multiplied each step (geometric, exponential growth or decay). You set up a recurrence relation, find any term with a closed formula, sum terms when needed, and apply all of this to financial contexts such as simple interest and reducing-balance depreciation. This dot point is Unit 3 Topic 2 and underpins the loans and annuities work in Unit 4.

The answer

Recurrence relations

A first-order recurrence relation defines each term from the previous one, together with a starting value. The two core forms are

t_{n+1} = t_n + d \quad \text{(arithmetic)}, \qquad t_{n+1} = r \, t_n \quad \text{(geometric)},

with a stated first term $t_1$ . Here $d$ is the common difference and $r$ is the common ratio.

Arithmetic sequences: linear change

An arithmetic sequence adds a constant $d$ each step, producing linear (straight-line) growth if $d > 0$ or decay if $d < 0$ . The $n$ th term is

t_n = t_1 + (n - 1) d.

The sum of the first $n$ terms is

S_n = \frac{n}{2} \bigl( 2 t_1 + (n - 1) d \bigr).

Simple interest is the classic arithmetic model: the same interest amount is added each period, so the balance grows linearly.

Geometric sequences: exponential change

A geometric sequence multiplies by a constant ratio $r$ each step, producing exponential growth if $r > 1$ or decay if $0 < r < 1$ . The $n$ th term is

t_n = t_1 \, r^{\,n-1}.

The sum of the first $n$ terms (for $r \ne 1$ ) is

S_n = \frac{t_1 (r^n - 1)}{r - 1}.

Reducing-balance depreciation is a geometric decay model: each year the asset keeps a constant fraction of its value, so $r = 1 - \text{depreciation rate}$ .

Choosing the right model

The key decision is whether the quantity changes by a constant amount (arithmetic) or a constant percentage (geometric). Constant dollars per year is arithmetic; constant percent per year is geometric. Reading this correctly is where most marks are won or lost.

Worked example

Simple interest versus reducing-balance depreciation

Part A: Simple interest (arithmetic). $\$ 2000 $earns simple interest of$ \ $120$ per year. The balance each year forms an arithmetic sequence with $t_1 = 2000$ (start) and $d = 120$ .

Recurrence: $t_{n+1} = t_n + 120$ , $t_1 = 2000$ .

Balance after $5$ years of interest:

t = 2000 + 5 \times 120 = 2000 + 600 = 2600.

Verify: $5 \times 120 = 600$ and $2000 + 600 = 2600$ . The balance is $\$ 2600$.

Part B: Reducing-balance depreciation (geometric). A $\$ 30,000 $machine depreciates at$ 20 $percent per year, so it retains$ 80 $percent:$ r = 0.80 $,$ t_1 = 30000$.

Recurrence: $t_{n+1} = 0.80 \, t_n$ , $t_1 = 30000$ .

Value after $3$ years:

t_4 = 30000 \times (0.80)^3.

Compute $(0.80)^3 = 0.512$ , so

t_4 = 30000 \times 0.512 = 15360.

Verify: $0.8 \times 0.8 = 0.64$ , $0.64 \times 0.8 = 0.512$ , and $30000 \times 0.512 = 15360$ . The machine is worth $\$ 15,360$ after three years.

Common traps

Mixing up arithmetic and geometric: Constant dollars per period is arithmetic (add $d$ ); constant percent per period is geometric (multiply by $r$ ). Choosing the wrong one breaks every later calculation.
Off-by-one in the index: The $n$ th term uses $(n - 1)$ , not $n$ , as the exponent or multiplier. The first term corresponds to $n = 1$ , so after $k$ steps you are at term $k + 1$ .
Using the wrong ratio for depreciation: If an asset depreciates at $20$ percent, the ratio is $r = 0.80$ , not $0.20$ . The ratio is the fraction retained.
Treating compound contexts as arithmetic: Compound interest multiplies each period and is geometric. Only simple interest is arithmetic.
Forgetting the starting term in a recurrence: A recurrence relation is incomplete without its first term; stating only $t_{n+1} = r t_n$ does not define the sequence.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 QCAA4 marksThe number of songs on a person's playlist, n, in each week since joining a music streaming service, t, forms an arithmetic sequence, as shown by the graph. The graph passes through (1, 87), (2, 209), (3, 331), (4, 453) and (5, 575). Use the arithmetic sequence to predict the number of songs on this person's playlist 25 weeks after joining the streaming service.

Show worked answer →

Step 1 - first term (1 mark): t_1 = 87 (the value at week 1).

Step 2 - common difference (1 mark): d = t_2 - t_1 = 209 - 87 = 122 (and 331 - 209 = 122 confirms it).

Step 3 - set up the nth-term rule (1 mark): t_n = t_1 + (n - 1)d = 87 + 122(n - 1).

Step 4 - evaluate at week 25 (1 mark): t_25 = 87 + 122 x 24 = 87 + 2928 = 3015 songs.

The standard trap is using n = 24 instead of n = 25, or forgetting the (n - 1). Week 1 is the first term, so 25 weeks after joining is the 25th term.

2022 QCAA7 marksThe first three lines in a pattern have the equations given. Their slopes form the terms of one sequence and their y-intercepts form the terms of another sequence. Each sequence is either arithmetic or geometric. Line 1: y = -0.8x + 1.2. Line 2: y = 0.4x + 2.7. Line 3: y = -0.2x + 4.2. Determine the coordinates of the point where Line 5 in the pattern intersects Line 1.

Show worked answer →

Step 1 - slope sequence (1 mark). The slopes -0.8, 0.4, -0.2 form a geometric sequence with t_1 = -0.8 and common ratio r = 0.4 / -0.8 = -0.5, so slope_n = -0.8 x (-0.5)^(n-1).

Step 2 - intercept sequence (1 mark). The intercepts 1.2, 2.7, 4.2 form an arithmetic sequence with t_1 = 1.2 and d = 1.5, so intercept_n = 1.2 + (n - 1) x 1.5.

Step 3 - slope of Line 5 (1 mark): m = -0.8 x (-0.5)^4 = -0.8 x 0.0625 = -0.05.

Step 4 - intercept of Line 5 (1 mark): c = 1.2 + 4 x 1.5 = 7.2, so Line 5 is y = -0.05x + 7.2.

Step 5 - x-coordinate of intersection (1 mark): set Line 1 = Line 5: -0.8x + 1.2 = -0.05x + 7.2, so -0.75x = 6, giving x = -8.

Step 6 - y-coordinate (1 mark): substitute into Line 1: y = -0.8 x (-8) + 1.2 = 6.4 + 1.2 = 7.6.

Step 7 - communicate clearly (1 mark): the intersection point is (-8, 7.6). The seventh mark rewards a logical, well-organised solution showing each sequence and the simultaneous solve.

2021 QCAA6 marksThe table shows the total number of times a new song is played on a music service in the days following its first release. Number of days since first release: 5, 10, 15, 20. Total number of times played ('000s): 8, 12, 18, 27. The songwriter is paid 0.175 cents every time their song is played and will be paid after 60 days. They predict that by then they will be owed at least $1000. Given that the number of times the song is played is increasing exponentially, evaluate the reasonableness of this prediction.

Show worked answer →

Step 1 - fit an exponential model (2 marks). Because plays grow exponentially, fit y = a x b^x to the four data points using exponential regression (equivalently, linearise by fitting log(y) against x). This gives approximately y = 5.33 x 1.0845^x, where y is in thousands and x is the number of days.

Step 2 - predict total plays at day 60 (1 mark): y = 5.33 x 1.0845^60 = 691.98 thousand, so about 691 980 plays.

Step 3 - convert plays to pay (1 mark): pay = 691 980 x 0.175 cents = 121 096 cents = $1210.97.

Step 4 - compare and evaluate (1 mark): $1210.97 is greater than$ 1000.

Step 5 - communicate the judgement (1 mark): the prediction is reasonable, because the exponential model forecasts about $1211 after 60 days, which exceeds the$ 1000 the songwriter expected.

Marks here reward choosing the exponential (geometric) model rather than a straight line, then carrying the units carefully from cents to dollars.